Question

Let z be a complex number such that the imaginary part of z is nonzero and $a=z^2+z+1$ is real. Then $a$ cannot take the value:

• A. $-1$
• B. $\frac{1}{3}$
• C. $\frac{1}{2}$
• D. $\frac{3}{4}$

Answer 1

Answer: (D)

$\frac{3}{4}$

The given equation is $z^2+z+1-a=0$

If the solution is not real then $\Delta =b^2-4ac$ of the quadratic $a{x}^2+bx+c=0$ is less than zero.

$\Rightarrow 1-4(1-a)<0$

$\Rightarrow 1-4+4a<0$

$\Rightarrow 4a<3$

$\Rightarrow a<\frac{3}{4}$

Hence, all values of a less than $\frac{3}{4}$ will give non-real solutions.

Options A, B and C are less than $\frac{3}{4}$.

Hence option D is correct